Question

我正在尝试以管理员身份登录。我使用了多个身份验证，我的用户工作正常，第二个用于管理，我正在尝试登录，但是当我在数据库中插入正确的值时，我手动插入了使用播种的值会给我一个错误，我不知道原因。

我所做的是配置 auth.php 所以我可以为用户和管理员进行两次身份验证。

<?php

return [



'defaults' => [
    'guard' => 'web',
    'passwords' => 'users',
],



'guards' => [
    'web' => [
        'driver' => 'session',
        'provider' => 'users',
    ],

    'api' => [
        'driver' => 'token',
        'provider' => 'users',
    ],

    'admins' => [
        'driver'   => 'session',
        'provider' => 'admins'
    ]
],



'providers' => [
    'users' => [
        'driver' => 'eloquent',
        'model' => App\Models\User::class,
    ],

     // For admin
    'admins' => [
        'driver' => 'eloquent',
        'model' => App\Models\Admin::class
    ]

    // 'users' => [
    //     'driver' => 'database',
    //     'table' => 'users',
    // ],
],



'passwords' => [
    'users' => [
        'provider' => 'users',
        'email' => 'auth.emails.password',
        'table' => 'password_resets',
        'expire' => 60,
    ],

    'admins' => [
        'provider' => 'admins',
        'email' => 'auth.emails.password',
        'table' => 'password_resets',
        'expire' => 60,
    ],
],

];

create_admins_table - 迁移

<?php

use Illuminate\Database\Schema\Blueprint;
use Illuminate\Database\Migrations\Migration;

class CreateAdminsTable extends Migration
{
/**
 * Run the migrations.
 *
 * @return void
 */
public function up()
{
    Schema::create('admins', function (Blueprint $table)
    {
        $table->increments('id');
        $table->string('username');
        $table->string('password');
        $table->timestamps();
    });
}
}

我的迁移文件已设置完毕。所以我迁移了它。

管理员 - 型号

<?php

namespace App\Models;

use Illuminate\Auth\Authenticatable;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Contracts\Auth\Authenticatable as AuthenticatableContract;

class Admin extends Model implements AuthenticatableContract
{
use Authenticatable;


protected $guard = "admins";

protected $table = 'admins';

protected $fillable = [
    'username',
    'password',
];

protected $hidden = [
    'password',
    'remember_token'
];
}

AdminTableSeeder - 种子

<?php

use Illuminate\Database\Seeder;

class AdminTableSeeder extends Seeder
{
public function run()
{
    DB::table('admins')->delete();

    $admins = array(
    array('username' => 'admin',
          'password' => Hash::make('admin'))
    );

    DB::table('admins')->insert($admins);
}
}

AdminController - 控制器

<?php

namespace App\Http\Controllers;

use Auth;
use App\Models\Admin;
use Illuminate\Http\Request;

class AdminController extends Controller
{
public function postAdminLogin(Request $request)
{
    $this->validate($request, 
    [
        'username' => 'required',
        'password' => 'required|min:7',
    ]);

    if(!Auth::attempt($request->only(['username','password']), $request->has('remember')))
    {
        dd('error');
    }

    dd('Succesful!');
}
}

更新!!

所以我将用户名和密码更改为letmein以避免混淆。我尝试登录正确的值，但我仍然无法登录。 !Auth:attempt与我的数据库中的值不匹配，并打印出“错误”消息。

public function postAdminLogin(Request $request)
{
    $this->validate($request, 
    [
        'username' => 'required',
        'password' => 'required|min:7',
    ]);

    if(!Auth::attempt($request->only(['username','password']), $request->has('remember')))
    {
        dd('error');
    }

    dd('Succesful!');
}

admin.blade.php

<form class = "form-vertical" role = "form" method = "post" action = "{{ route ('auth.admin') }}">

            <div class = "form-group {{ $errors->has('username') ? 'has-errors' : '' }}">
                <label for = "username" class = "control-label">Username</label>
                <input type = "text" name = "username" class = "form-control">

                @if ($errors->has('username'))
                    <span class = "help-block">{{ $errors->first('username') }}</span>
                @endif
            </div>

            <div class = "form-group {{ $errors->has('password') ? 'has-errors' : '' }}">
                <label for = "password" class = "control-label">Password</label>
                <input type = "password" name = "password" class = "form-control">

                @if ($errors->has('password'))
                    <span class = "help-block">{{ $errors->first('password') }}</span>
                @endif

            </div>

            <div class = "checkbox">
                <label>
                    <input type = "checkbox" name = "remember">Remember me</input>
                </label>
            </div>

            <div class = "form-group">
                <button type = "submit" class = "btn btn-default">Login</button>
            </div>

            <input type = "hidden" name = "_token" value = "{{ Session::token() }}">

        </form>

Answer 1

你的表名在这里是问题...我们认为它应该是AdminTableSeeder中的“admins”，如下所示：

        use Illuminate\Database\Seeder;

        class AdminTableSeeder extends Seeder {

            public function run() {
                DB::table('admins')->delete();  //<== "admins" AS DEFINED IN YOUR CONFIG::: NOT "admin"

                $admin = array(
                array('username' => 'admin',       
                      'password' => Hash::make('admin'))
                );

                DB::table('admins')->insert($admin);
            }
        }

Answer 2

AdminTableSeeder - 种子

使用truncate而不是delete

use Illuminate\Database\Seeder;

class AdminTableSeeder extends Seeder
{
public function run()
{
  DB::table('admins')->truncate();

  $admins = array(
  array('username' => 'admin',
        'password' => Hash::make('admin'))
  );

  DB::table('admins')->insert($admins);
 }
 }

Answer 3

错误消息旁边的SQL查询显示

select count(*) as aggregate from `admin` where `password`=admin

现在，在您的WHERE子句中，密码被重新添加，因此我假设它是一个列（不是表）。反引号通常在SQL中用于表名或列名。但是你将列名称与admin进行比较，没有刻度。＆＃39;管理＆＃39;不是保留字，并且在查询中不会引用它作为文本字符串，因此数据库服务器开始查找具有该名称的表以匹配。

那么，你试过在那里添加引号吗？

SELECT COUNT(*) AS aggregate FROM `admin` WHERE `password`='admin'

编辑：

如果所有其他方法都失败了，您可能还想简单地转储正在发布的数据库查询。添加一个监听器/转储：

\DB::listen(function($sql, $bindings, $time) {
    var_dump($sql);
    var_dump($bindings);
    var_dump($time);
});

..并检查输出。

SQLSTATE [42S02]：找不到基表或视图：1146

3 个答案: